Let $P\subset\Bbb R^d$ be a convex polytope (convex hull of finitely many points). A $k$-in-sphere of $P$ is a sphere centered at the origin to which each $k$-face of $P$ is tangent. So a 0-in-sphere contains all the vertices and is actually a circumsphere, and a $(d-1)$-in-sphere is completely contained in $P$.


Question: If $P$ has $k$-in-spheres for all $k\in\{0,...,d-1\}$, is $P$ a regular polytope?

By definition, all these spheres are centered at the origin, hence are concentric.

The answer to the question is Yes for polygons. For $d\ge 3$ note that this property of $P$ is inherited by its faces, and it follows that all 2-faces of $P$ are regular polygons and all edges are of the same length.

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    $\begingroup$ Do you assume that the spheres are concentric, or do you say that they must be concentric? $\endgroup$ – Wlodek Kuperberg Jun 1 at 19:46
  • $\begingroup$ @WlodekKuperberg I assume this. Otherwise e.g. any triangle (not necessarily regular) would be an example as well. $\endgroup$ – M. Winter Jun 1 at 19:48

This is true in all dimensions, and can be proved by induction (on $d$) applied to the following (slightly stronger) hypothesis:

Theorem: If $P$ is a convex $d$-polytope with $k$-in-spheres for all $k \in [0, d-1]$, then:

  • $P$ is regular.
  • $P$ is determined (up to an element of the orthogonal group $O(d)$) by that $d$-tuple $(r_0, r_1, \dots, r_{d-1})$ of $k$-in-radii.
  • $P$ is determined completely if additionally a facet (codimension-1 face) $Q$ of $P$ is specified.

Proof: If the polytope $P$ has squared $k$-in-radii $(r_0^2, r_1^2, \dots, r_{d-1}^2)$, then every facet of $P$ has squared $k$-in-radii $(r_0^2 - r_{d-1}^2, r_1^2 - r_{d-1}^2, \dots, r_{d-2}^2 - r_{d-1}^2)$. By the first two parts of the inductive hypothesis, all facets of $P$ are therefore regular and congruent to each other (being determined by these $k$-in-radii).

Now, given a facet $Q$ of $P$ and a facet $R$ of $Q$, let $\Pi$ be the hyperplane through the origin which contains $R$. Let $Q'$ be the other facet of $P$ which contains $R$. Because the $k$-in-spheres of $Q'$ are the reflections (in $\Pi$) of the $k$-in-spheres of $Q$, and they share a common facet $R$, it follows (from the third part of the inductive hypothesis) that $Q'$ is the reflection of $Q$ through the hyperplane $\Pi$.

As the boundary $\partial P$ (i.e. the union of all facets) is homeomorphic to $S^{d-1}$, we can reach any facet $Q_1$ from any facet $Q_0$ by a 'path' of 'adjacent' (i.e. sharing a common subfacet) facets. Consequently, we can transform any facet into any other facet by a sequence of reflections in hyperplanes through the origin. As each facet is flag-transitive, it therefore follows that $P$ is flag-transitive (i.e. regular) as desired.

Moreover, this reflection procedure of building $P$ from a single facet $Q$ establishes the third part of the theorem.

This leaves the second part of the theorem. Suppose $P$ and $P'$ are two polytopes sharing the same set of $k$-in-spheres. Let $Q$ be an arbitrary facet of $P$, and $Q'$ be an arbitrary facet of $P'$. By the inductive hypothesis, $Q$ and $Q'$ are congruent; let $f$ be an isometry of the ambient space which maps $Q$ to $Q'$. The origin is either mapped to itself or (if we chose the 'wrong' isometry) to $2v$, where $v$ is the centroid of $Q$; we can if necessary reflect again in the hyperplane containing $Q$ to ensure the origin is preserved by $f$. Consequently, $f$ is an element of the orthogonal group $O(d)$ which maps $Q$ to $Q'$. By the third part of the theorem (which we've already proved), $f$ must map $P$ to $P'$, establishing the second part of the theorem.

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In $R^3$, since the spheres are concentric, not only all faces are regular, but also all edges are of the same length, and all faces are inscribed in circles of the same radius, hence are congruent. Also, all dihedral angles between faces with a common edge are equal, which implies that all vertices are of the same valence. This makes the polytope regular. It seems that this reasoning can be generalized to all dimensions.

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  • $\begingroup$ How exactly you show that "all faces are inscribed in circles of the same radius" (I know that they are inscribed, but why same radius)? Also I would be interested in how exactly this generalizes. $\endgroup$ – M. Winter Jun 1 at 21:51
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    $\begingroup$ Each face lies on a plane which is tangent to the insphere, and intersects the circumsphere in a circle (namely the circumcircle of that face). The squared radius of this circumcircle is the difference between the squared radius of the circumsphere and the squared radius of the insphere (by Pythagoras). $\endgroup$ – Adam P. Goucher Jun 1 at 21:55
  • $\begingroup$ @M.Winter: All centers of these circles are at the same distance from the origin, and all circles lie on one sphere, the one containing the vertices. $\endgroup$ – Wlodek Kuperberg Jun 1 at 22:01
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    $\begingroup$ I apologise for my wrong edit. I actually read it carefully and thought I knew what you meant, but I should have asked rather than changing it when I realised I wasn't sure. $\endgroup$ – LSpice Jun 1 at 22:41
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    $\begingroup$ @LSpice -- Thanks, no harm done. $\endgroup$ – Wlodek Kuperberg Jun 1 at 23:19

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